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 November 17th, 2015, 10:35 PM #1 Senior Member     Joined: Jan 2012 Posts: 723 Thanks: 7 Solve the equation completely. Solve the equation completely. $$4^{(2x+3/2)}-65(4^x)+8=0$$ How do I tackle this? Last edited by Chikis; November 17th, 2015 at 10:55 PM.
 November 17th, 2015, 11:42 PM #2 Senior Member     Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18 4^(2x + 3/2) - 65(4^x) + 8 = 0 ((4^2x) * 4^(3/2)) - 65(4^x) + 8 = 0 [[(4^x)^2] * 8] - 65(4^x) + 8 = 0 let 4^x = a... Then the expression becomes... 8a^2 -65a + 8 = 0 8a^2 - 64a - a + 8 = 0 8a(a - 8 ) -1(a - 8 ) = 0 (8a - 1)(a - 8 ) = 0 so, a = 1/8 or a = 8 so, a = 4^x = 1/8 or 4^x = 8 4^x = 2^2x = 2^-3 or 2^2x = 2^3 so... 2x = -3 or 2x = 3 x = -3/2 or x = 3/2
November 18th, 2015, 03:32 AM   #3
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Quote:
 Originally Posted by Chikis How do I tackle this?
Factor out the $4^\frac32 = 8$ in the first term to leave a quadratic in $4^x$.

November 18th, 2015, 04:10 AM   #4
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Quote:
 Originally Posted by Chikis Solve the equation completely.
Will the real solutions suffice?

November 18th, 2015, 07:48 AM   #5
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Quote:
 Originally Posted by shunya 4^(2x + 3/2) - 65(4^x) + 8 = 0 ((4^2x) * 4^(3/2)) - 65(4^x) + 8 = 0 $\displaystyle \ \ \ \$ * You need "4^(2x)" instead. [[(4^x)^2] * 8] - 65(4^x) + 8 = 0 let 4^x = a... Then the expression becomes... 8a^2 -65a + 8 = 0 8a^2 - 64a - a + 8 = 0 8a(a - 8 ) -1(a - 8 ) = 0 (8a - 1)(a - 8 ) = 0 so, a = 1/8 or a = 8 so, a = 4^x = 1/8 or 4^x = 8 4^x = 2^2x = 2^-3 or 2^2x = 2^3 $\displaystyle \ \ \ \$** You need "2^(2x)" in two places instead. so... 2x = -3 or 2x = 3 x = -3/2 or x = 3/2

* $\displaystyle \ \ \ \$4^2x = (4^2)x $\displaystyle \ \ \ \$You don't mean that.

** $\displaystyle \ \$2^2x = (2^2)x $\displaystyle \ \ \ \$You don't mean that.

 November 30th, 2015, 06:55 PM #6 Senior Member     Joined: Jan 2012 Posts: 723 Thanks: 7 Thanks from Chikis.

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